I'm working on this equilibrium homework problem and I'll post all my work. Can you help me please? I'm a little stuck especially on the polynomial.
If iodine vapor and bromine vapor are reacted in 15o degree C they form idodine monobromide gas. At 150 degrees C the value of Kc is known to be 125. Lets say we placed 5.73 g of I2 and 4.91 g of Br2 into a 5.00L flask and heat it to 150 degrees C.
find the value of the final equilibrium concentrations of all reactants and products of M?
so far this is what i have:
When converting I2 to mols I got 0.0226 mols and divided by 5.00L gives me 0.00452 M.
For Br 2 i got 0.0307 mol and divided by 5.00 L gives me 0.00614 M.
(Q=0 b/c no products exist initially)
i would prefer to use as simple math as possible, so i didn't use multiple variables.
let 2x=IBr
x=I2
x=Br2
so the expression from that is:
2x^2/(0.00452-x)(0.00614-x)=125
when expanded, for the polynomial, i calculated
4x^2/(2.76X10^-5 - 0.00452x-0.00641x - .00345)
4x^2 = 125x^2-.7675X-.565x-.00345
0 = 121x^2 - .7675x-.565x-.00345
so i solved for x using the quadratic formula and i received
-1.3325+- square root(-1.3325)^2-4(125)(.000345)/ 2(121)
-1.3325+-.3253/242
x=-.00685 and -.50365 that can't be right can it?
negative values wouldn't work for the domain correct?
Let's go through the steps to solve this equilibrium problem.
Step 1: Convert the masses of I2 and Br2 to moles.
You correctly calculated the number of moles of I2 and Br2:
- Moles of I2 = 0.0226 mol
- Moles of Br2 = 0.0307 mol
Step 2: Calculate the initial concentrations.
Since the reaction takes place in a 5.00 L flask, you can divide the moles by the volume to get molar concentrations:
- Initial concentration of I2 = 0.0226 mol / 5.00 L = 0.00452 M
- Initial concentration of Br2 = 0.0307 mol / 5.00 L = 0.00614 M
Step 3: Write the expression for the equilibrium constant, Kc.
The balanced chemical equation for the reaction is:
I2(g) + Br2(g) ⇌ IBr(g)
The expression for Kc is obtained from the coefficients of the balanced equation:
Kc = [IBr]^2 / ([I2][Br2])
Step 4: Set up the equation using the expression for Kc.
Let's denote x as the change in concentration of I2, Br2, and IBr at equilibrium.
Then, the final equilibrium concentrations can be expressed as follows:
- Concentration of I2 at equilibrium = 0.00452 - x
- Concentration of Br2 at equilibrium = 0.00614 - x
- Concentration of IBr at equilibrium = 2x
Now, substitute these concentrations into the Kc expression:
125 = (2x)^2 / ((0.00452 - x)(0.00614 - x))
Step 5: Solve the equation.
To solve for x, we need to expand the equation and rearrange it to the standard form of a quadratic equation. However, it seems there was an error in your polynomial expansion. Let's correct it:
125 = (4x^2) / ((0.00452 - x)(0.00614 - x))
Multiply both sides by (0.00452 - x)(0.00614 - x):
125(0.00452 - x)(0.00614 - x) = 4x^2
Expand and rearrange the equation:
121x^2 - 0.7743x + 0.000258268 = 4x^2
117x^2 - 0.7743x + 0.000258268 = 0
Now, we can solve this quadratic equation to find the values of x.
Using the quadratic formula:
x = (-(-0.7743) ± √((-0.7743)^2 - 4(117)(0.000258268))) / (2(117))
Calculating this equation should give you the correct values for x, which represent the change in concentration at equilibrium.